JEE Main 2020 Question Paper with Solution (9th January - Evening)
74 Questions MCQ Test Additional Study Material for JEE | JEE Main 2020 Question Paper with Solution (9th January - Evening)
| 1 Crore+ students have signed up on EduRev. Have you? |
A mass m attached to spring of natural length
and spring constant k.One end of string is attached to centre of disc in horizontal plane which is being rotated by constant angular speed ω. Find extension per unit length in spring (given k >> mω2) :
and spring constant k.One end of string is attached to centre of disc in horizontal plane which is being rotated by constant angular speed ω. Find extension per unit length in spring (given k >> mω2) :
A loop of radius R and mass m is placed in a uniform magnetic field B with its plane perpendicular to the field. Current I is flowing in it. Now loop is slightly rotated about its diameter and released. Find time period of oscillation.
A string of mass per unit length μ = 6 × 10–3 kg/m is fixed at both ends under the tension 540 N. If the string is in resonance with consecutive frequencies 420 Hz and 490 Hz. Then find the length of the string?
Fundamental frequency = 490 – 420
= 70 Hz
Ratio of energy density of two steel rods is 1 : 4 when same mass is suspended from the rods. If length of both rods is same then ratio of diameter of rods will be.
stress x 
A particle is projected from the ground with speed u at angle 60° from horizontal. It collides with a second particle of same mass moving with horizontal speed u in same direction at highest point of its trajectory. If collision is perfectly inelastic then find horizontal distance travelled by them after collision when they reached at ground
so horizontal range after collision
H-like atom with ionization energy of 9R. Find the wavelength of light emitted (in nm) when electron jumps from second excited state to ground state. (R is Rydberg constant)
wavelength = 
nm
λ = 11.39 nm
Two planets of masses M and M/2 have radii R and R/2 respectively. If ratio of escape velocities from their surfaces v1/v2 is n/4, then find n :
Find centre of mass of given rod of linear mass density λ = 
, x is distance from one of its end. Length of the rod is l.
If a point source is placed at a depth h in a liquid of refractive index 4/3. Find percentage of energy of light that escapes from liquid. (assuming 100% transmission of emerging light)
Solid angle dΩ = 2pR2 (1 - cos β)
Percentage of light = 
System is released from rest. Moment of inertia of pulley 'I'. Find angular speed of pulley when m1 block falls by 'h'. (Given m1 > m2 and assume no slipping between string and pulley).
k1+ U1 = k1 + k2
Find the current supplied by the battery?
Both diodes are in reverse biased
An AC source is connected to the LC series circuit with V = 10 sin (314t). Find the current in the circuit as function of time ? (L = 40 mH, C = 100 μF)
There is a long solenoid of radius ‘R’ having ‘n’ turns per unit length with current i flowing in it. A particle having charge ‘q’ and mass ‘m’ is projected with speed 'v' in the perpendicular direction of axis from a Point on its axis Find maximum value of 'v' so that it will not collide with the solenoid.
A Capacitor C and resister R are connected to a battery of 5V in series. Now battery is disconnected and a diode is connected as shown in figure (a) and (b) respectively. Then charge on the capacitor after time RC in (a) and (b) respectively is QA and QB. Their value are
Maximum charge on capacitor = 5CV
(a) is forward biased and
(b) is reverse biased for case (a)
A sphere of density ρ is half submerged in a liquid of density σ and surface tension T. The sphere remains in equilibrium. Find radius of the sphere (assume the force due to surface tension acts tangentially to surface of sphere)
An EM wave is travelling in 
direction. Axis of polarization of EM wave is found to be 
Then equation of magnetic field will be
EM wave is in direction 
Electric field is in direction ---> kˆ
--> direction of propagation of EM wave
Different value of a, b and c are given and their sum is d. Arrange the value of d in increasing order
Two gases Ar (40) and Xe (131) at same temperature have same number density. Their diameters are 0.07 nm and 0.10 nm respectively. Find the ratio of their mean free time
Mean free time = 
A particle starts moving from origin with velocity
from origin and acceleration 
. Find x-coordinate at the instant when y-coordinate of the particle is 32 m.
t = 4 sec
= 60 m.
An electron 
is released in Electric field E from rest. Rate of change of de-Broglie wavelength with time will be.
In YDSE pattern with light of wavelength λ1 = 500nm, 15 fringes are obtained on a certain segment of screen. If number of fringes for light of wavelength λ2 on same segment of screen is 10, then the value of λ2 (in nm) is-
If in a meter bridge experiment, the balancing length ' l ' was 25 cm for the situation shown in the figure. If the length and diameter of the of wire of resistance R is made half, then find the new balancing length in centimetre is
l = 40.00 cm
Find the power loss in each diode (in mW), if potential drop across the zener diode is 8V.
Power loss in each diode = (4)(10–2) W = 40 mW
An ideal gas at initial temperature 300 K is compressed adiabatically 
of its initial volume. The gas is then expanded isobarically to double its volume. Then final temperature of gas round to nearest integer is:
PVy = constant
TVy–1 = constant
TB = 300 × 28/5
Now for BC process
Tc = 1819 K
If electric field in the space is given by 
and electric flux through ABCD is φ1 and electric flux through BCEF is φ2 , then find (φ1 - φ2)
Flux via ABCD
Flux via BCEF
5 g of Zn reacts with
(I) Excess of NaOH (II) Dilute HCl, then volume ratio of H2 gas evolved in (I) and (II) is
Zn + 2NaOH ---> Na2ZnO2 + H2
Zn + 2HCl ---> ZnCl2 + H2
According to stoichiometry in both the reactions, equal number of moles of H2 are evolved.
Given Ksp for Cr(OH)3 is 6 × 10–31 then determine [OH–].
(Neglect the contribution of OH– ions from H2O)
Select the correct statements among the followings
(A) LiCl does not dissolve in pyridine
(B) Li does not react ethyne to form ethynide.
(C) Li and Mg react slowly with water.
(D) Among alkali metals Li has highest hydration tendency.
Theory based
Given an element having following ionisation enthalpies IE1 = 496 
and IE2 = 4562 
one mole
hydroxide of this element is treated separated with HCl and H2SO4 respectiv ely. Moles of HCl and H2SO4 reacted respectively is
According to the given data of I.E, This element must belong to group 1 and thus is monovalent & form hydroxide of the type M(OH).
MOH + HCl --> MCl + H2O
1 mole 1 mole
2MOH + H2SO4 ---> M2SO4 + H2O
1 mole 1/2 mole
Reactant A represented by square is in equilibrium with product B represented by circles. Then value of equilibrium constant is
Given following complexes
(I) Na4[Fe(CN)6]
(II) [Cr(H2O)6] Cl2
(III) (NEt4)2 [CoCl4]
(IV) Na3[Fe(C2O4)3] (Δ0 > P)
Correct order of spin only magnetic moment for the above complexes is.
Select the correct option :
Entropy is function of temperature and also entropy change is function of temperature.
ΔS = ∫dq/T
ST = ∫0T ncdT/T
A compound (A ; B3N3H3Cl3) reacts with LiBH4 to form inorganic benzene (B). (A) reacts with (C) to form B3N3H3(CH3)3. (B) and (C) are respectively.
In a box a mixture containing H2, O2 and CO along with charcoal is present then variation of pressure with the time will be as follows :
Theory based.
Given complex [Co(NH3)4Cl2]. In it if Cl – Co – Cl bond angle is 90º then it is :
Amongst the following which has minimum conductivity.
Theory based.
Number of sp2 hybrid orbitals in Benzene is :
In benzene total six sp2 hybrid carbon atoms are present. Each carbon atom has 3 sp2 hybrid orbitals.
Therefore total sp2 hybrid orbitals are 18 in benzene.
Which of the following reaction will not form racemic mixture as product?
(CH3)2CH–CH=CH2 --HCl----> 
---Hybride shift---> 
In which compound C–Cl bond length is shortest?
Resonance form of Cl–CH=CH–NO2 is more stable than resonance form of any other given compounds. Hence, double bond character in carbon-chlorine bond is maximum and bond length is shortest.
Biochemical oxygen demand (BOD) is defined as ............ in ppm of O2.
The amount of oxygen required by bacteria to break down the organic matter present in a certain volume of a sample of water, is called Biochemical Oxygen Demand (BOD).
Monomer(s) of which of the given polymer is chiral?
In PHBV, both monomers have chiral centre.
Which of the following options is correct ?
The order of basic character is :
(I) 
(II) 
(III) 
(IV) 
Basic strength depends upon availability of lone pairs. Greater the resonance of lone pairs lesser the basic strength.
Compound A 
Compound A will be :
Compound X
Compound X will be :
Total number of Cr–O bonds in Chromate ion and dichromate ion is.
Chromate ion ---> 
Dichromate ion ---> 
=> Total number of Cr and O bonds is 12.
Lacto bacillus has generation time 60 min. at 300 K and 40 min. at 400 K. Determine activation energy in 
(given wrong in paper)
ln(3/2) × 8.3 × 1200 = Ea
=> Ea = 0.4 × 8.3 × 1200
=> Ea = 3984 J/mol.
=> Ea = 3.984 kJ/mol.
One litre sea water (d = 1.03g/cm3) contains 10.3 mg O2 gas. Determine concentration of O2 in ppm.
0.1 ml of an ideal gas has volume 1 dm3 in a locked box with friction less piston. The gas is in thermal equilibrium with excess of 0.5 m aqueous ethylene glycol at its freezing point. If piston is released all of a sudden at 1 atm then determine the final volume of gas in dm3 (R = 0.08 atm L mol–1 K–1 Kf = 2.0 K molal–1).
Kf = 2.0
m = 0.5 m
ΔTf = Kfm = 0.5 × 2
Tinitial = 272 K
n = 0.1 mol
V = 1 dm3
= 2.176 atm
After releasing piston P1V1 = P2V2
2.176 × 1 = 1 × V2
V2 = 2.176 dm3
Compound A 
Percentage carbon in compound A is:
Compound A is 
Percentage carbon in compound A = 
If f(x) = 
g(x) = 
then find the area bounded by f(x) and g(x) from x = 
.
Required area = Area of trapezium ABCD – 
z is a complex number such that |Re(z)| + |Im (z)| = 4 then |z| can't be
z = x + iy
|x| + |y| = 4
Minimum value of |z| = 2√2
Maximum value of |z| = 4
So |z| can't be √7
If f(x) = 
and a – 2b + c = 1 then
Apply R1 = R1 + R3 – 2R2
f(x) = 1
f(50) = 1
Let an is a positive term of a GP and 
Let GP is a, ar, ar2 ……..
= a3 + a5 + ….. a201 = 200
............(1)
= a2 + a4 + ……. a200 = 100
........... (2)
Form (1) and (2) r = 2
add both
a2 + a3 + ……….. a200 + a201 = 300
r(a1 + ………… a200) = 300
= 300/r = 150
If 
y(1) = 1 and y(x) = e then x = ?
Put y = vx
When x = 1, y = 1 then 
So x = √3e
Let probability distribution is
then value of p(x > 2) is
6k2 + 5k = 1
6k2 + 5k – 1 = 0
6k2 + 6k – k – 1 = 0
(6k – 1) (k + 1) = 0
k = – 1 (rejected); k = 1/6
P(x > 2) = k + 2k + 5k2
= λtanq + 2log f(x) + c then ordered pair (λ, f(x)) is
tanθ = t => sec2θ dθ = dt
= – t + 2 log (1+t) + C
= –tanθ + 2 log (1 + tanθ) + C
= λ = -1 and f(x) = 1 + tanθ
If p ---> (p ∧ ~ q) is false. Truth value of p & q will be
If 
= A then the value of x at which f(x) = [x2] sinπx is discontinuous.
(where [.] denotes greatest integer function)
4 – 0 = A
check when
(A) x = 
=> x = √5 => discontinuous
(B) x = A + 21 => x = 5
=> continuous
(C) x = A => x = 2
=> continuous
Let one end of focal chord of parabola y2 = 8x is 
, then equation of tangent at other end of this focal chord is
Let 
is (2t2, 4t)
=>
Parameter of other end of focal chord is 2
=> point is (8, 8)
=>equation of tangent is 8y – 4(x+8) = 0
=> 2y - x = 8
Let x + 6y = 8 is tangent to standard ellipse where minor axis is 
, then eccentricity of ellipse is
, 
Equation of tangent ≡ y = mx ± 
comparing with 
and 
if f(x) and g(x) are continuous functions, fog is identity function, g'(b) = 5 and g(b) = a then f'(a) is
If 7x + 6y – 2z = 0
3x + 4y + 2z = 0
x – 2y – 6z = 0 then which option is correct
= 7(–20) – 6(–20) – 2(–10)
= – 140 + 120 + 20 = 0
so infinite non-trivial solution exist
now equation (1) + 3 equation (3)
10x – 20z = 0
x = 2z
Let x = 2sinθ - sin2θ and
y = 2 cosθ - cos2θ
find 
at θ = π
then correct choice is
F'(x) = x2g(x)
=> F'(1) = 1.g(1) = 0 ....(1)
(∴ g(1) = 0)
F''(1) = 0 + 1 × 3
=> F''(1) = 3 .....(2)
From (1) and (2) F(x) has local minimum at x = 1
Let both root of equation ax2 – 2bx + 5 = 0 are α and root of equation x2 - 2bx - 10 = 0 are α and β. Find the value of α2 + β2
b2 = 5a ..... (i) (a ≠ 0)
α + β = 2b ..... (ii)
and αβ = – 10 ......(iii)
α = b/a is also root of x2 - 2bx - 10 = 0
b2 – 2ab2 – 10a2 = 0
by (i) 5a – 10a2 – 10a2 = 0
20a2 = 5a
a = 1/4 and b2 = 5/4
α2 = 20 and β2 = 5
Now α2 + β2
= 5 + 20
= 25
Let A = {x : |x| < 2} and B = {x : |x – 2| ≥ 3} then
Let x = 
and y = 
where θ ∈ (0,π/4), then
y = 1 + cos2θ + cos4θ + .......
= > y (1 – x) = 1
Let the distance between plane passing through lines 
and 
and plane 23x – 10y – 2z + 48 = 0 is 
then k is equal to
Lines must be intersecting
(2s – 1, 3s + 3, 8s –1)
= (2t – 3, t – 2 , λt + 1)
2s - 1 = 2t - 3,
3s + 3 = t - 2 ,
8s - 1 = λt + 1
=> t = -1, s = -2 , λ = 18
distance of plane contains given lines from given plane is same as distance between point (–3, –2,1) from given plane.
Required distance equal to
k = 3
If 25C0 + 5 25C1 + 9 25C2 ……… 101 25C25 = 225 k find k = ?
So k = 51
Let circles (x – 0)2 + (y – 4)2 = k and (x – 3)2 + (y – 0)2 = 12 touches each other than find the maximum value of 'k'
Two circles touches each other if C1C2 = = |r1 ± r2|
Distance between C2(3, 0) and C1(0, 4) is either √k + 1 or 
(C1C2 = 5)
=> √k + 1 = 5 or 
=> k = 16 or k = 36
=> maximum value of k is 36
Let 
angle between 
equal to π/3
If
is perpendicular to 
then find the value of 
Also 
Number of common terms in both sequence 3, 7, 11, ………….407 and 2, 9, 16, ……..905 is
First common term = 23
common difference = 7 × 4 = 28
Last term ≤ 407
=> 23 + (n-1) × 28 ≤ 407
=> (n - 1) × 28 ≤ 384
=> n ≤ 13.71 + 1
n ≤ 14.71
So n = 14
If minimum value of term free from x for 
is L1 in θ ∈ 
and L2 in θ ∈ 
find 
∵ {Min value of L1 at θ = π/4}
= 16C8 . 28.24 {∵ min value of L2 at θ = π/8]
|
22 videos|162 docs|17 tests
|
|
Use Code STAYHOME200 and get INR 200 additional OFF
|
Use Coupon Code |
|
22 videos|162 docs|17 tests
|
Download free EduRev App
Related content with JEE Main 2020 Question Paper with Solution (9th January - Evening)
|
|
|
|
Related tests with JEE Main 2020 Question Paper with Solution (9th January - Evening)
|
|
|
|
|
|
© EduRev
|
Education Revolution
|
Follow Us
|
